Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ac}{1+c} \geq 3.$ 
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1+ab}{1+a}+\dfrac{1+bc}{1+b}+\dfrac{1+ac}{1+c} \geq 3.$$

This looks symmetric, so should I prove it for just $a \leq b \leq c$ and then the other cases follow?
 A: Since $1=abc$, we have
\begin{align}
\dfrac{1+ab}{1+a} + \dfrac{1+bc}{1+b} + \dfrac{1+ca}{1+c} & = \dfrac{abc+ab}{1+a} + \dfrac{abc+bc}{1+b} + \dfrac{abc+ca}{1+c}\\
& = ab \dfrac{1+c}{1+a} + bc \dfrac{1+a}{1+b} + ca \dfrac{1+b}{1+c}\\
& \geq 3\sqrt[3]{ ab \dfrac{\color{red}{1+c}}{\color{blue}{1+a}} \cdot  bc \dfrac{\color{blue}{1+a}}{\color{orange}{1+b}} \cdot  ca \dfrac{\color{orange}{1+b}}{\color{red}{1+c}}}\,\,\,\, \left(\because\text{AM-GM}\right)\\
& = 3\sqrt[3]{(abc)^2}\\
& = 3
\end{align}
A: By the AGM inequality $$(1/3) \sum (1+a b)/(1+a)\geq (P/Q)^{1/3}$$ $$\text {where }\;  P=\prod (1+a b)\quad  \text {and } Q=\prod (1+a).$$ $$\text {Now}\quad P=\prod ((c+a b c )/c)=\prod (c+1)/\prod c=\prod (c+1)=Q.$$
A: Another way.
Let $a=\frac{y}{x},$ $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Thus,  $c=\frac{x}{z}$ and  by AM-GM we obtain: $$\sum_{cyc}\frac{1+ab}{1+a}=\sum_{cyc}\frac{1+\frac{z}{x}}{1+\frac{y}{x}}=\sum_{cyc}\frac{x+z}{x+y}\geq3\sqrt[3]{\prod_{cyc}\frac{x+z}{x+y}}=3.$$
